Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:20000007")
#pragma GCC optimize("unroll-loops")
const int N = (int)1e6 + 100;
const int maxn = (int)1e3 + 100;
const int base = (int)1e9;
const int mod = (int)1e9 + 7;
const int inf = INT_MAX;
const long long ll_inf = LLONG_MAX;
const long double pi = acos((long double)-1.0);
const long double eps = 1e-8;
void solve() {
  int n;
  cin >> n;
  map<int, bool> is;
  for (int i = 1; i * (i + 1) <= 2 * (int)1e9; ++i) is[i * (i + 1)] = true;
  for (int i = 1; i * (i + 1) <= 2 * (int)1e9; ++i) {
    int j = 2 * n - i * (i + 1);
    int j_sqrt = sqrt(j);
    if (is[j]) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  int T = 1;
  for (; T; --T) solve();
  return 0;
}